f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)当方程f(x)+a-1=0有解时,求a的取值范围
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/17 19:26:31
f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)当方程f(x)+a-1=0有解时,求a的取值范围f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)当方程f(x)+a-1=0有
f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)当方程f(x)+a-1=0有解时,求a的取值范围
f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)当方程f(x)+a-1=0有解时,求a的取值范围
f(x)=2sin(2x+2A+π/6)+1,A为常数,(1)当方程f(x)+a-1=0有解时,求a的取值范围
-1<=sin(2x+2A+π/6)<=1
所以-1<=2sin(2x+2A+π/6)+1<=3
f(x)=-a+1
所以有解的-1<=-a+1<=3
-3<=a-1<=1
-2<=a<=2
f(x)-m=2sin(2x-π/3) 1-m=2 2sin(2x-π/3)=m 2-1=m 1 x∈【π/4,π/2】 2x-π/3∈【π/6,π/2】 1≤2sin(2x-π/3)≤2
f(x)+a-1=0
即2sin(2x+2A+π/6)+1+a-1=2sin(2x+2A+π/6)+a=0 有解
即a=-2sin(2x+2A+π/6) 所以a∈【-2,2】
函数f(x)=sin(2x+a) -π
函数f(x)=sin(2x+a)(-π
f(x)=sin(2x+π/6)怎么求导
已知函数f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a的最大值为1最后化简为f(x)=2sin(x+π/6)+a 我想问最大值1指的是f(x)还是sin(x+π/6)如果值得是f(x) 那 sin(x+π/6)怎么算
已知F(X)=根号3COS^2 X+SIN XCOS X-2SIN X*SIN(X-π/6),求F(X)的最大值
若f(x)=sin πx/6,则f(1)+f(2)+...+f(102)=?
求f(x)=sin(2x+a)的奇偶性
已知函数f(x)=2sin(2x+π/6)+a+1.
已知函数f x=a(2sin ²x/2+sin x)+b
设f(x)=sin(2x+π/6)+3/2,g(x)=f(x+a).若g(x)为偶函数,求a的值.
若函数f(x)=sin(x+a)为偶函数,则f(π/2)=
设关于X的函数f(x)=sin(2x+a)(-π
设关于X的函数f(x)=sin(2x+a)(-π
f(x)=(6sin^4x-7sin^2x+2)/(sin^2x-cos^2x)
f(sin^2x)=x/sinx 求f(x)
已知函数f(x)=sin(2x+π/6)+2sin^2(x+π/6)-2cos^2x+a-1化简
f(x)=sin(x+π/6)+sin(x-π/6)+cosx+af(x)=2sinxcos(π/6)+cosx+a用的是什么公式?
函数f(x)=cos2x+cos(x+π/3)+sin(x+π/6)+3sin^2x的最小值A.0 B.2 C.9/4 D.3